Does this function belong to an interesting class of functions?

AI Thread Summary
The function f(φ) = (1 - φ) / (φ - 1) simplifies to -1 except at φ = 1, where it is undefined due to a discontinuity. The discussion explores whether this function belongs to a broader class of functions with unique characteristics or applications, noting its similarity to Mobius transformations, primarily used in complex analysis. There is debate over the interpretation of the function's value at φ = 1, with some suggesting it could be defined as 1 in specific contexts, though this contradicts its established definition. The distinction between different function definitions is emphasized, particularly in relation to practical applications in physics. Overall, the conversation highlights the complexities and nuances of defining mathematical functions and their implications.
PhilDSP
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Hello and thanks for your consideration,

I'd like some insight into the function f(\phi) = \frac {1 - \phi}{\phi - 1}

Does this apply to any known modeling situations? Is it recognized as belonging to a more general class of functions that may have interesting or unique characteristics? Or can the function be transformed into a function that does?
 
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Your function, as written, is just equal to -1, except when \phi = 1, where there is a discontinuity because the denominator vanishes there.

If you want an example of a function that has a similar form but isn't trivially some constant and has some applications, see Mobius transformation. (But note that the Mobius transformation is usually used with complex numbers. I don't know if it is used much in real number applications).
 
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Thanks, an association with the Mobius transformation does yield many interesting things to think about, especially since \phi can be complex in the situation where the function popped up.

We could argue that the value becomes 1 when \phi = 1 couldn't we? This almost sounds like a spinning sphere where the axis must be aligned parallel to a force acting on the sphere, but which can suddenly undergo a spin flip.
 
PhilDSP said:
We could argue that the value becomes 1 when \phi = 1 couldn't we?

No, you can't define a function one way and then argue that it has a different definition. You can, however, define a function that is 1 when \phi = 1 and equal to -1 elsewhere. You can argue that this definition applies to a certain practical situation. That would be an argument about physics.
 
PhilDSP said:
We could argue that the value becomes 1 when \phi = 1 couldn't we?

No. "f(x)= -1" and "g(x)= -1 if x\ne 1, and is not defined at x= 1" are two different functions.
 
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